\documentclass[12pt]{article} 

\topmargin=-0.1in 
\textheight=8.5in 

\begin{document}

\noindent{\bf HW \#3, 269C } {\bf \ \ \ Due on Monday, May 10} 

\noindent{\bf [1]} Show that the Lax-Milgram theorem contains as particular case the 
Riesz representation theorem (in other words, assume Lax-Milgram and prove the Riesz representation theorem). 

\medbreak

\noindent{\bf [2]} Let $V$ be a Hilbert space, and $a:V\times V\rightarrow R$ be a bilinear form. Show that: $a$ bounded is equivalent with $a$ continuous. 

\medbreak 

\noindent{\bf [3]} Let $V$ be a Hilbert space and the (nonlinear) operator $A:V\rightarrow V$, satisfying 

(i) there is $M\geq 0$ s.t. $\forall$ $u,v\in V$, $\|Au-Av\|\leq M\|u-v\|$ 

(ii) there is $\alpha >0$ s.t. $\forall$ $u,v\in V$, $\langle Au-Av,u-v\rangle \geq \alpha \|u-v\|^2$ 

Show that the nonlinear equation $Au=f$ has a unique solution (for $f\in V$), using the Banach fixed point theorem and the same technique for proving the Lax-Milgram theorem (introduce the function $g_{\lambda}$). 

\medbreak 

\noindent{\bf [4]} Let $V$ be a complex Hilbert space, $a:V\times V\rightarrow C$ a sesquilinear form, $L:V\rightarrow C$ an anti-linear form. Assume that $a$ is Hermitian, thus $a(v,u)=\overline{a(u,v)}$, $\forall u,v\in V$, and that $a(v,v)\geq 0$. Consider the problems 

$$\mbox{(V)} \mbox{ Find } u\in V \mbox{ s.t. } a(u,v)=L(v), \ \forall v\in V,$$

$$\mbox{(M)}  \mbox{ Find } u\in V \mbox{ s.t. } J(u)=\inf_{v\in V}J(v),$$
with $J:V\rightarrow R$ defined by $J(v)=\frac{1}{2}a(v,v)-\mbox{Re}L(v)$. 

Show that $u\in V$ is solution of (V) iff $u\in V$ is solution of (M). 

\medbreak 

\noindent{\bf [5]} Prove Poincar\'e inequality on $H^1_0(0,1)$. 


\bigbreak 

%\noindent{\bf [6]} Let $\Omega=\{x\in R^2:\ |x|\leq 1\}$. Show that the function $v(x)=|x|^{\alpha}$ belongs to $H^1(\Omega)$ if $\alpha>0$. 

%\bigbreak

\noindent{\bf [6]} Consider the problem with an inhomogeneous boundary condition, 
$$
\left\{
\begin{array}{l} 
-\triangle u=f\mbox{ in } \Omega,\\
u=u_0 \mbox{ on } \Gamma=\partial\Omega,
\end{array}
\right.
$$
where $f$ and $u_0$ are given. Show that this problem can be given the 
following equivalent variational formulations: 

$(V)$ Find $u\in V(u_0)$ such that $a(u,v)=(f,v)$, $\forall v\in H^1_0(\Omega)$,

$(M)$ Find $u\in V(u_0)$ such that $F(u)\leq F(v)$, $\forall v\in V(u_0)$,

where $V(u_0)=\{v\in H^1(\Omega): \ v=u_0 \mbox{ on } \Gamma\}$. 

Then formulate a finite element method and prove an error estimate (as in Thm. 1.1, page 24). 

Recall: $H_0^1(\Omega)=\{v\in L^2(\Omega), \ \nabla v\in L^2(\Omega)^n,\ v=0\mbox{ on }\partial\Omega\}$, where $n$ is the spatial dimension. 

\bigbreak


{\bf Useful notes} 

$\bullet$ Over the complex space $C$, with the above notations, we say that $a:V\times V\rightarrow C$ is a is sesquilinear form if $a$ linear in the first argument (under the usual addition and scalar multiplication), $a$ antilinear in the second argument (usual addition but $a(u,\lambda v)=\bar{\lambda}a(u,v)$). Note that if $\mbox{Re} a(v,v)\geq \alpha \|v\|^2$, together with $a$ and $L$ bounded, then the Lax-Milgarm Lemma holds in the complex case with the same proof. 

$\bullet$ {\bf Thm.} Let $v\in H^1(a,b)$ and denote by $v'\in L^2(a,b)$ its first order distributional derivative. Then for almost every $x,y\in (a,b)$, we have $$v(y)-v(x)=\int_{y}^x v'(t)dt.$$





\end{document}









